Solution

Drift velocity is achieved when particle's path is unaffected respect to force fields that acting over it. So, absolute values of these forces are equal and their directions are opposed. Hence,

$$mg = q~v_d~B$$

$$v_d = \frac{mg}{qB}$$

Now, considering the following data, calculations for each particle are possible: mass of electron $m_e = 9.1 \cdot 10^{-31}\text{ kg}$, mass of proton $m_p = 1.672 \cdot 10^{-27}\text{ kg}$, absolute value of charge $q = 1.6 \cdot 10^{-19}\text{ C}$ and acceleration due to gravity $g = 9.8 \text{ m/s}^2$. Calculating,

$$v_e = 7.9625 \cdot 10^{-7} \text{ m/s}$$

$$v_p = 1.423625 \cdot 10^{-3} \text{ m/s}$$